B and C are absurd; if a series converges, it must have a sum, but if a series diverges, it cannot have a sum.
Now, notice that
![\frac12+\frac29+\frac4{27}+\frac8{81}+\cdots=\frac12+(2^(2-1))/(3^2)+(2^(3-1))/(3^3)+(2^(4-1))/(3^4)+\cdots](https://img.qammunity.org/2019/formulas/mathematics/middle-school/lcdvxapf81z6moypk1s0j8b2cpkox63xqf.png)
That is, we can write the sum more compactly as
![\frac12+\frac12\displaystyle\sum_(n=1)^\infty\left(\frac23\right)^n](https://img.qammunity.org/2019/formulas/mathematics/middle-school/2eutppzkxgbiflr778p24pyn7bzgrpz18r.png)
The series is geometric with common ratio
![\frac23<1](https://img.qammunity.org/2019/formulas/mathematics/middle-school/2wrfq5nfi5h49mnmr1fleuefym7dy0zugf.png)
, so the series converges (and thereby has a sum), so the answer is D.